Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 Å. Assuming density of the oxide as `4.0gcm^(-3)`, then the number of `Fe^(2+)andO^(2-)` ions present in each unit cell will be

Options

A. Four `Fe^(2+) and` four `O^(2-)`
B. Two `Fe^(2+)` and four `O^(2-)`
C. Four `Fe^(2+)` and two `O^(2-)`
D. Three `Fe^(2+)` and three `O^(2-)`

Answer

A

Explanation

Let the units of ferrous oxide in a unit cell=n, mol. Wt. of ferrous oxide=56+6=72 g `mol^(-1)`
Wt. of n units=`(72xxn)/(6.023xx10^(23))`, Vol. of one unit cell=`("length of edge")^(3)=(5A^(@))^(3)=125xx10^(-24)cm^(3)`
Density=`("wt. of cell")/("volume"),4.0=(72xxn)/(6.023xx10^(23)xx125xx10^(-24))`
`n=(3011.5xx10^(-1))/(72)=41.8xx10^(-1)=4.18~~4`.